A linear transformation that maps $(1,2,3)$ to $(3,2,1)$ and vice versa.

Question

Question: Let $v_1=(1,2,3)^T$ and $v_2=(3,2,1)^T$. Find a standard matrix for a linear transformation that maps $v_{1}$ to $v_{2}$ and vice versa.

I’m interested in different ways of going about this problem.

I offer one way.

One way: Let $A=[a_1 a_2 a_3]$ with $a_i \in \mathbb{R^3}$. Then this leads to a system with vectors as unknowns,

$$a_1+2a_2+3a_3=v_2$$

$$3a_1+2a_2+a_3=v_1$$

Now we use “augmented form” to make calculations easier, and reduce it to “RREF”, and get our solution.

$$\begin{bmatrix} 1 & 0 & -1 & \frac{v_1-v_2}{2} \\ 0 & 1 & 2 & \frac{3v_2-v_1}{4} \end{bmatrix}$$

For example take Let $a_3=0$, this leads to:

$$A=\begin{bmatrix} -1 & 2 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$

Answer 1

Another way is to find $v_3$ such that $\{v_1,v_2,v_3\}$ forms a basis for $\mathbb R^3$ (e.g. $v_3=(1,0,0)^T$). Then one can take $A$ to be $\pmatrix{0&1&0\\1&0&0\\0&0&0}$ under this basis. And then transform $A$ to the standard basis.

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Hope this helps.

Answer 2

You may use

$$A=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$

which is a permutation matrix.

This is easily obtained by swapping the first and the third row of the identity matrix.

Answer 3

Following from one of the comments we can use a permutation matrix in this scenario, and conjugate it. There is already an answer involving permutation matrices that is shorter, and in some sense simpler. I think my solution is a little bit more general, since it would work for any two vectors you want to swap.

For instance, consider the linear map $T_{12}$ that sends $e_1 \to e_2$, $e_2 \to e_1$ and $e_3 \to e_3$. This would have have matrix

$$ \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Now, let $S$ be the linear maps that sends $e_1 \to v_1$, $e_2 \to v_2$, and $e_3 \to e_3$. This would, of course, have matrix:

$$ \begin{bmatrix}1 & 3 & 0 \\ 2 & 2 & 0 \\ 3 & 1 & 1 \end{bmatrix}$$

If you compute it, the determinant of this matrix is $-4$. Look at the composition: $S \circ T_{12} \circ S^{-1}$.

Acting on $v_1$, we have:

$$S \circ T_{12} \circ S^{-1}(v_1) = S \circ T(e_1) = S(e_2) = v_2$$

and acting on $v_2$, we have:

$$S \circ T_{12} \circ S^{-1} (v_2) = S \circ T_{12}(e_2) = S(e_1) = v_1$$

And there you have it. This map also fixes $e_3$. The matrix for $S^{-1}$ is $$\begin{bmatrix} -\frac{1}{2} & \frac{3}{4} & 0 \\ \frac{1}{2} & -\frac{1}{4} & 0 \\ 1 & -2 & 1 \end{bmatrix}$$

The matrix for this map will be:

$$ \begin{bmatrix}1 & 3 & 0 \\ 2 & 2 & 0 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{3}{4} & 0 \\ \frac{1}{2} & -\frac{1}{4} & 0 \\ 1 & -2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 0 \\0 & 1 & 0 \\2 & -2 & 1 \end{bmatrix} $$